# Reverse Pairs Problem

## Description

LeetCode Problem 493.

Given an integer array nums, return the number of reverse pairs in the array.

A reverse pair is a pair (i, j) where 0 <= i < j < nums.length and nums[i] > 2 * nums[j].

Example 1:

``````1
2
Input: nums = [1,3,2,3,1]
Output: 2
``````

Example 2:

``````1
2
Input: nums = [2,4,3,5,1]
Output: 3
``````

Constraints:

• 1 <= nums.length <= 5 * 10^4
• -2^31 <= nums[i] <= 2^31 - 1

## Sample C++ Code

``````1
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class Solution {
private:
int count;

void checkCount(vector<int>& nums, int start, int mid, int end) {
// two pointers;
int l = start, r = mid + 1;
while (l <= mid && r <= end) {
if ((long)nums[l] > (long) 2 * nums[r]) {
count += (mid - l + 1);
r++;
} else {
l++;
}
}
// worst case might be nlog(n)
sort(nums.begin() + start, nums.begin() + end + 1);
return;
}

void mergeSort(vector<int>& nums, int start, int end) {
if (start == end) return;

int mid = (start + end)/2;
mergeSort(nums,start, mid);
mergeSort(nums,mid+1,end);

checkCount(nums,start,mid,end);
return;
}

public:
int reversePairs(vector<int>& nums) {
if (!nums.size()) return 0;
count = 0;
mergeSort(nums,0,nums.size()-1);
return count;
}
};
``````